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The perimeter of the rectangle below is 130 units. Find the value of z

aev [14]

Answer:

9 =z

Step-by-step explanation:

The perimeter of a rectangle is

P = 2(l+w) where l is the length and w is the width

P = 2(3z+2+ 4z)

Combine like terms

P = 2(7z+2)

Substitute in the perimeter

130 = 2( 7z+2)

Divide each side by 2

130/2 = 2/2 ( 7z+2)

65 = 7z+2

Subtract 2 from each side

65-2 = 7z+2-2

63 = 7z

Divide by 7

63/7 = 7z/7

9 =z

8 0

3 years ago

Read 2 more answers

Your friend believes that he has found a route to work that would make your commute faster than what it currently is under simil

Strike441 [17]

Answer:

The range of p-values

0.01

Step-by-step explanation:

Explanation:-

Given random sample size 'n' = 7

Assume that the populations are normally distributed

Null Hypothesis :H₀:μd=0.

Alternative Hypothesis:H₁:μd<0.

Degrees of freedom

ν = n-1 =7-1 =6

given the test statistic t = - 3.201

we will use single tailed test in t-distribution table

The test statistic t= 3.201 is lies between the critical values is 0.01 and 0.025

The range of p-values

0.01

Final answer:-

The range of p-values

0.01

5 0

3 years ago

Find the equation in stope intercept form of the line that passes through the points (-2,4) and (6,2),

Juli2301 [7.4K]

Answer:

y = -x/4 + 7/2

Step-by-step explanation:

Find the slope

( -2 , 4) ( 6 , 2)

m = (y2 - y1)/(x2 - x1)

x1 = -2

y1 = 4

x2 = 6

y2 = 2

m = ( 2 - 4)/( 6 -(-2)

m = ( 2 -4)/(6 +2)

= -2/8

= -1/4

Using point slope form equation

y - y1 = m( x - x1)

Using the second point

( 6 , 2)

x1= 6

y1=2

m = -1/4

y - 2 = -1/4(x - 6)

y - 2 = -(x - 6)/4

Open the bracket with -

y - 2 = (-x + 6)/4

y = ( -x + 6)/4 + 2

LCM = 4

y =( -x + 6 + 8)/4

y = (-x + 14)/4

Rearrange using slope intercept form

y = mx + c.

y = -x/4 + 14/4

We can break 14/4 by dividing by 2

y = -x/4 + 7/2

8 0

4 years ago

Select the correct answer.

katen-ka-za [31]

Answer:

d

Step-by-step explanation: bc ik

7 0

3 years ago

A rectangle is three centimeters.longer than it is wide. if its length were to be decreased by two centimeters, its area would d

Scorpion4ik [409]

Given/equations formed from text:
I: l=w+3
II: A=l*w
III: new area B=(l-2)*w
IV: B=A-30

substitute l in II with I to remove l:
II': A=l*w=
(w+3)*w=
w^2+3w

substitute l in III with I to remove l:
III': B=(l-2)*w=
(w+3-2)*w=
(w+1)*w=
w^2+w

substitute A and B from II' and III' into IV:
B=A-30
w^2+w=w^2+3w-30
30=2w
w=15

insert w=15 into I:
l=15+3=18
->
A) dimensions of full size rectangle are width 15 and length 18
dimensions of reduced size rectangle are width 15 and length 16

B) full size: 15*18=10*18+5*18=180+90=270cm^2
reduced size: 15*16=10*16+5*16=160+80=240cm^2

7 0

4 years ago